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The 4th, 6th and 9th term of an arithmetical progression form the first three times of a geometric progression. If the first term of the arithmetic progression is 3 determine the common difference of the arithmetic progression and common ratio of the geometric al progression

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4 answers
  1. 4 th ==> 3+3d = a
    6 th ==> 3 + 5d = a r
    9 th ==> 3 + 8 d = a r^2
    --------------------------------------
    15 + 15 d = 5 a
    9 + 15 d = 3 a r
    -------------------------- subtract
    6 = a (5-3r)
    ---------------------------
    24 + 40 d = 8 a r
    15 + 40 d = 5 a r^2
    ----------------------------- subtract
    9 = a ( 8 r - 5 r^2)
    so
    9 / ( 8 r - 5 r^2) = 6 / (5-3r)
    9 (5 - 3 r) = 6 (8 r - 5 r^2)
    45 - 27 r = 48 r - 30 r^2
    30 r^2 - 75 r + 45 = 0

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  2. I believe the roots are r = 1.5 and r = 1
    try those

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  3. if r = 1
    6 = a (5-3r)
    6 = a (2)
    a = 3
    a, a, a ------- boring, forget it
    if r = 1.5
    6 = a (5-3r) = 5 a - 4.5 a = .5 a
    a = 12
    12 , 18 , 27 first three geometric
    3+3d = a
    3 d = 12 - 3 = 9
    d = 3
    3,6,9,12 YES !!! the fourth arith is the first geo

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  4. from the given:
    (a+5d)/(a+3d) = (a+8s)/(a+5d)
    but a = 3
    (3+5d)^2 = (3+3d)(3+8d)
    9 + 30d + 25d^2 = 9 + 33d + 24d^2
    d^2 - 3d = 0
    d(d-3) = 0
    d = 0, or d = 3

    so for the AP, when d = 3
    term(4) = 3 + 3(3) =12
    term(6) = 3 + 5(3) = 18
    term(9) = 3 + 8(3) = 27

    sure enough: 18/12 = 3/2, and 27/18 = 3/2, so they are also in a GP
    so r = 3/2

    if d = 0,
    term(4), term(6), and term(9) are all equal to 3
    so 3,3,3 technichally would be a GP with r = 1

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